Henry_2001 wrote: Let $a,b,c$ be the lengths of the sides of a given triangle.If positive reals $x,y,z$ satisfy $x+y+z=1,$ find the maximum of $axy+byz+czx.$ ***********************************************Solution 1************************************************************************** *********************************************************************************************************************************** Let's consider the triangle with side lengths $\left(\sqrt{a}, \sqrt{b}, \sqrt{c}\right)$. Applying Kooi's inequality, we get: $$(x+y+z)^2R^2 \geq axy+byz+czx$$On the other hand, we have: $$R^2=\left(\frac{\sqrt{abc}}{4S}\right)^2=\frac{abc}{2(ab+bc+ca)-a^2-b^2-c^2}$$Therefore, we have: $$axy+byz+czx \leq \frac{abc}{2(ab+bc+ca)-a^2-b^2-c^2 }$$Equality at: $$(x,y,z)=\left(\frac{b(c+a-b)}{2(ab+bc+ca)-a^2-b^2-c^2 },\frac{c(a+b-c)}{2(ab+bc+ca)-a^2-b^2-c^2 },\frac{a(b+c-a)}{2(ab+bc+ca)-a^2-b^2-c^2 }\right)$$ ***********************************************Solution 2************************************************************************** *********************************************************************************************************************************** Using Lagrange Multipliers, we can prove that: $$axy+byz+czx \leq \frac{abc}{2(ab+bc+ca)-a^2-b^2-c^2 }$$With equality at: $$(x,y,z)=\left(\frac{b(c+a-b)}{2(ab+bc+ca)-a^2-b^2-c^2 },\frac{c(a+b-c)}{2(ab+bc+ca)-a^2-b^2-c^2 },\frac{a(b+c-a)}{2(ab+bc+ca)-a^2-b^2-c^2 }\right)$$which verifies the condition: $$x+y+z = \frac{\sum_{cyc}a(b+c-a)}{2(ab+bc+ca)-a^2-b^2-c^2 } = \frac{2(ab+bc+ca)-a^2-b^2-c^2 }{2(ab+bc+ca)-a^2-b^2-c^2 }=1$$

The intended solution is likely to consider a triangle $\triangle{ABC}$ with $BC=\sqrt{a}$, $CA=\sqrt{b}$, and $AB=\sqrt{c}$. Then, $-axy-byz-cxz$ is the power of a point from a point with barycentric coordinates $P=(x,y,z)$ to the circumcircle of $ABC$. Note that if the circumradius is $R$ and the circumcenter is $O$, then $$axy+byz+czx=R^2-PO^2 \geq R^2=(\frac{\sqrt{abc}}{4[ABC]})^2=\frac{abc}{2ab+2bc+2ca-a^2-b^2-c^2}$$. Now, we show that equality is attainable. The condition is equivalent to $P$ being inside of $\triangle{ABC}$, and $a+b > c \Rightarrow BC^2+CA^2 \geq AB^2$ (and similar statements), so $\triangle{ABC}$ is acute. Thus, $O$ is inside of $\triangle{ABC}$, so equality is attainable, as desired.

mira74 wrote: The intended solution is likely to consider a triangle $\triangle{ABC}$ with $BC=\sqrt{a}$, $CA=\sqrt{b}$, and $AB=\sqrt{c}$. Then, $-axy-byz-cxz$ is the power of a point from a point with barycentric coordinates $P=(x,y,z)$ to the circumcircle of $ABC$. Note that if the circumradius is $R$ and the circumcenter is $O$, then $$axy+byz+czx=R^2-PO^2 \geq R^2=(\frac{\sqrt{abc}}{4[ABC]})^2=\frac{abc}{2ab+2bc+2ca-a^2-b^2-c^2}$$. Now, we show that equality is attainable. The condition is equivalent to $P$ being inside of $\triangle{ABC}$, and $a+b > c \Rightarrow BC^2+CA^2 \geq AB^2$ (and similar statements), so $\triangle{ABC}$ is acute. Thus, $O$ is inside of $\triangle{ABC}$, so equality is attainable, as desired. $BC=a,CA=b,AB=c,$ according to the original(Chinese) version.

Henry_2001 wrote: Let $a,b,c$ be the lengths of the sides of a given triangle.If positive reals $x,y,z$ satisfy $x+y+z=1,$ find the maximum of $axy+byz+czx.$ https://artofproblemsolving.com/community/c6h1096588p4917744

Nguyenhuyen_AG wrote: Henry_2001 wrote: Let $a,b,c$ be the lengths of the sides of a given triangle.If positive reals $x,y,z$ satisfy $x+y+z=1,$ find the maximum of $axy+byz+czx.$ https://artofproblemsolving.com/community/c6h1096588p4917744 Well, I notice that these two problems have something similar. However, I fail to solve this problem with yours. Could you explain the link between these two problems?

Henry_2001 wrote: Nguyenhuyen_AG wrote: Henry_2001 wrote: Let $a,b,c$ be the lengths of the sides of a given triangle.If positive reals $x,y,z$ satisfy $x+y+z=1,$ find the maximum of $axy+byz+czx.$ https://artofproblemsolving.com/community/c6h1096588p4917744 Well, I notice that these two problems have something similar. However, I fail to solve this problem with yours. Could you explain the link between these two problems? It is the same problem...

@Moved to post 2

This problem can be also solved using Klamkin's inequality...

Henry_2001 wrote: $BC=a,CA=b,AB=c,$ according to the original(Chinese) version. I know that $a,b,c$ satisfy the triangle inequality, but i chose to define a triangle with sides $\sqrt{a},\sqrt{b},\sqrt{c}$ instead to use this approach.

bel.jad5 wrote: Let's consider the triangle with side lengths $\left(\sqrt{a}, \sqrt{b}, \sqrt{c}\right)$. Applying Kooi's inequality Are you sure that's the name of the inequality? I couldn't find it on the net.

Godfather2043 wrote: bel.jad5 wrote: Let's consider the triangle with side lengths $\left(\sqrt{a}, \sqrt{b}, \sqrt{c}\right)$. Applying Kooi's inequality Are you sure that's the name of the inequality? I couldn't find it on the net.

Please check the following link Kooi inequality

Also here: Kooi's inequality

It is also mentioned in my book, Theorem 79: Kindle version (Look Inside)

Henry_2001 wrote: mira74 wrote: The intended solution is likely to consider a triangle $\triangle{ABC}$ with $BC=\sqrt{a}$, $CA=\sqrt{b}$, and $AB=\sqrt{c}$. Then, $-axy-byz-cxz$ is the power of a point from a point with barycentric coordinates $P=(x,y,z)$ to the circumcircle of $ABC$. Note that if the circumradius is $R$ and the circumcenter is $O$, then $$axy+byz+czx=R^2-PO^2 \geq R^2=(\frac{\sqrt{abc}}{4[ABC]})^2=\frac{abc}{2ab+2bc+2ca-a^2-b^2-c^2}$$. Now, we show that equality is attainable. The condition is equivalent to $P$ being inside of $\triangle{ABC}$, and $a+b > c \Rightarrow BC^2+CA^2 \geq AB^2$ (and similar statements), so $\triangle{ABC}$ is acute. Thus, $O$ is inside of $\triangle{ABC}$, so equality is attainable, as desired. $BC=a,CA=b,AB=c,$ according to the original(Chinese) version. He considers another triangle in order to interpret the expression in terms of Barycentric coordinate.

Consider that $axy+byz+czx=axy+(by+cx)z$ $=axy+(by+cx)(1-x-y)$ $=axy+by+cx-by^2-cx^2-bxy-cxy$ $=-by^2+(b+(a-b-c)x)y-cx^2+cx$ $\leqslant(-cx^2+cx)-\frac{{(b+(a-b-c)x)}^2}{4(-b)}$ $=\frac{(a^2+b^2+c^2-2ab-2bc-2ca)x^2+2b(a-b+c)x+b^2}{4b}$ $\leqslant\frac{b^2-\frac{{(2b(a-b+c))}^2}{4(a^2+b^2+c^2-2ab-2bc-2ca)}}{4b}$ $=\frac{abc}{2ab+2bc+2ca-a^2-b^2-c^2}.$ $($The maximum of $f(x)=ax^2+bx+c$ is $c-\frac{b^2}{4a}$ $\leqslant0$$).$ Note : By Ravi substitution $a=u+v,b=v+w,c=w+u.$ Then $ a^2+b^2+c^2-2ab-2bc-2ca=-4(uv+vw+wu)\leqslant0. $